User wlodzimierz holsztynski - MathOverflow

Answer by Wlodzimierz Holsztynski for Why is Set, and not Rel, so ubiquitous in mathematics?

2013-05-20T06:23:38Z

It's hard to edit long multi-part Answers. Sorry for writing another one.

@Qfwfq: It's a common knowledge that for centuries perhaps there was a vague identification of function-formulas and functions. Even in the quite modern times many textbooks say that a function is a rule assigning a value to an argument--something in this spirit. They are shy about domains, and especially about the codomains.

Even as late as fifteen years ago, and perhaps today too, typical exercises from typical Calculus textbooks present students with formulas, and ask them to provide the domain. This implies that the domain is not a part of the definition, that it is something to be deduced. A tricky exercise would be for formula

$$\frac{x^2-1}{x+1}$$

On one hand, as it is now, $-1$ is forbidden (illegal); and on the other hand this function is--hm--equal to $x-1$. I don't know if there is a uniform policy among all these undergraduate Calculus textbooks.

After WWI the Cantor set theory certainly found its home in Poland where about each mathematical monograph had a nice introduction on set theory and elements of point-set topology, and they used a clean and consistent set-theoretical language. It had amounted to certain mathematical culture, which was modern in the pre-categorical times. Today this style, at a more advanced stage, is represented for instance by Engelking's monograph on General Topology. In particular, he makes a clear distinction between the Cartesian and diagonal product of functions (the latter being these days a standard notion of the theory of categories). For a contrast, the great Soviet mathematics, powerful and imposing as it was, was not so fond of such niceties. In particular, outstanding mathematician and great writer and teacher Igor Schafarevich would be at peace with calling diagonal product of mappings to be a cartesian product. His lectures were still great of course. Many excellent and useful Soviet mathematical books looked naive and old-fashioned from the point of view of set theoretical notation.

Nevertheless, even with the Polish set-theoretical tradition, the definition of a function from $X$ to $Y$ would sound something like: a set $f$ of ordered pairs $(x\ y)\in f$ such that:

$$\forall_{x\in X}\exists_{y\in Y}\quad(x\ y)\in f$$
$$\forall_{x\in X}\forall_{y\ y'\in Y}\quad\left(\left(\left(x\ y\right)\ \left(x\ y'\right)\ \in\ f\right)\ \ \Rightarrow\ \ \left(y=y'\right)\right)$$

This is pretty good, and better then a function is a rule but it implicitly creates not one but two notions: a function, and a function from $X$ to $Y$, the former being just a graph (I call them functional graphs or function graphs in general, i.e. a function graph is simply any set which has only ordered pairs for its elements; hm, graf would be a perfect name, with f indicating a connection with the notion of a function).

The ultimate explicit treatment of functions as ordered triples consisting of domain, codomain, and the graph, is due to the theory of categories, I would think.

REMARK It is ironic that in the case of SET alone--in an isolation from other categories via functors--it is not necessary to specify the domain (it is implied by the graph), while we still have to indicate the codomain.

@Qfwfq, I hope that the above rumbling partially answers your question. If you'd like me I may still check a few books in my possesion for specific references and quotes. (BTW, may I address you QfwfQ? Sometimes I like symmetry).

Answer by Wlodzimierz Holsztynski for Why is Set, and not Rel, so ubiquitous in mathematics?

2013-05-18T23:34:25Z

@Ronnie Brown:

The emphasis on (total) functions comes from the understanding reflected by theory of categories (e.g. by algebraic topology). In the past the distinction between a total function and a partial function was not accented. The functions induced by partial functions are so drastically different from the total case in algebraic topology that there arose the need of a very strong distinction between the partial and the total functions. Also for the more subtle distinction of two functions with the same graph, and the same domain but different codomains (most of the time one is contained in the other in the mathematical practice).

Thus today when we deal with a partial function we just call it a (total) function but with a smaller domain; thus we hardly see partial functions anymore.

@mbsq, Alice in Wonderland payed words to mean what she wanted them to mean. Mathematical community is Alice in Wonderland (except here and there for an old age dementia). Let $B\ \ S$ be the unit closed ball in $R^n$ and its boundary (sphere), so that $S\subseteq B\subseteq R^n$. Consider three "identity" functions, the later two induced by the first one:

$$ I_B:B\rightarrow B$$ $$ i_{SB} : S\rightarrow B$$ $$ I_S : S\rightarrow S$$

so that $i_{SB} = I_B|S$. In the old times (of mostly local analysis) the "identity" functions were kind of taken for granted and mostly invisible like well behaved children in strict families. There was not much need to talk about them explicitly. In that old spirit you are saying that to call $I_S$ and $i_{SB}$ different functions is absurdity. However, let $H_n$ be the n-th homology functor. The homomorphisms induced by the identity inclusions are no more "identities", i.e. they are not necessarily monomorphisms, they can be in general just arbitrary homomorphisms. Thus they cannot be treated like polite subdued children anymore. In particular we need in our notions and notation an easy distinction between $I_S$ and $i_{SB}$ because the induced homomorphisms are dramatically different: $H_n(I_S) = I_{\mathbb Z}$ is a non-zero identity homomorphism on $\mathbb Z$, while $H_n(i_{SB})$ is the zero homomorphism. When two objects induce so much different effects then it is not an absurdity to call these objects different. These days children are not subdued anymore. Our definitions should reflect the importance of the codomain (as $S$ versus $B$ in the example above).

Is there a better function (linear or even a projection)?

2013-05-03T03:35:01Z

Let $A$ be a finite $n$-element set. Let $\mathbb R^A$ be an $n$-dimensional Euclidean space (with the ordinary Euclidean distance). Let $X$ be an arbitrary topological space. Consider a continuous map $f : X\rightarrow \mathbb R^A$.

DEFINITION A point $y\in \mathbb R^A$ is called essential (with respect to $f$) $\Leftarrow:\Rightarrow$ there exists a real $\epsilon > 0$ such that for every continuous $g : X\rightarrow \mathbb R^A$ such that the uniform distance is small: $|g-f| < \epsilon$, point $y$ is a value of $g$, i.e. there exists $x := x_g\in X$ such that $g(x)=y$.

Now consider continuous maps $f:X\rightarrow \mathbb R^B$ $\phi : \mathbb R^B \rightarrow \mathbb R^A$, where $B$ is an $m$-element set, $|B| = m > n = |A|$, and such that the composition $\phi\circ f:X\rightarrow \mathbb R^A$ has an essential value.

QUESTIONS:

Does there exist a linear map $\lambda : \mathbb R^B\rightarrow \mathbb R^A$ such that $\lambda\circ f: X\rightarrow \mathbb R^A$ has an essential value?
Does there exist an $n$-element set $C\subset B$ such that $\pi^B_C \circ f:X\rightarrow\mathbb R^C$ has an essential value? (where $\pi^B_C:\mathbb R^B\rightarrow \mathbb R^C$ is the canonical projection).

Here is a special case ($\dim$ stands for the topological dimension): assume that $X\subseteq\mathbb R^B$, and that $\dim(X) \ge n$.

QUESTIONS:

Does there exist a linear map $\lambda : \mathbb R^B\rightarrow \mathbb R^A$ such that $\lambda|X: X\rightarrow \mathbb R^A$ has an essential value? (we continue to assume that $|A|=n < m$).
Does there exist an $n$-element set $C\subset B$ such that $\pi^B_C | X: X \rightarrow\mathbb R^C$ has an essential value?

Answer by Wlodzimierz Holsztynski for Why is Set, and not Rel, so ubiquitous in mathematics?

2013-05-18T23:05:12Z

Here are my impressions:

Are apples more important than fruits? (Usually we tend to compare apples and oranges :-).

We have--in the finite case--a numerical metaphor which indicates a vague answer. The number of functions $A\rightarrow B$ is $|B|^{|A|}$, while the number of the relations in $A\times B$ is:

$$2^{|A|\times |B|}\ \ =\ \ \left(2^{|B|}\right)^{|A|}$$

The number of relations is overwhelming. The number of functions is positioned between "too few" and "too many"; and the meaning of a function is more meaningful, it is between "too special" and "too general". The notion of a relation is by comparison "too general" to play intensively a leading role by itself, as such.

The notion of a (self)function $A\rightarrow A$ can be compared to other oranges like linear orderings (i.e. permutations), partial orderings and equivalence relations in $A$. Let $a:=|A|$. There are $a^a$ of self-functions, $a!$ of linear orderings, over $2^{\lfloor a/2\rfloor\cdot\lceil a/2\rceil}$ of partial orderings but still under $a!\cdot 2^{a\cdot(a-1)/2}$ (hm, a lot), and there are under $a^a$ of equivalence relations.

Functions represent longing for determinism (predictability).

Isn't the accent on functions relatively modern? Ancient Greeks didn't talk much about functions.

PS. The multitude of partial orders combined with the importance of the notion (they happen to be $T_0$-topologies too :-) is puzzling.

General and translational Birkhoff lattices. Equational classes.

2013-05-16T10:04:55Z

By lattice I'll mean Birkhoff lattice.

The two classical equational classes of lattices are modular lattices and distributive lattices. The old problem used to be:

Is there an equational class between the modular class and the distributive class (different from both)?

Is this question still open?

One could attack it by getting an intimate knowledge of a large family of lattices which would amount in the full classification of that family. The hope would be of finding an intermediate equational class by pointing to a member of the said large family.

First let's mention that $\mathbb R^n$ admits a partial order

$$ x\le y\quad\Leftarrow:\Rightarrow\quad \forall\ _{k=1}^n\ \ x_k\le y_k$$

for every $x\ y\in\mathbb R^n$. Space $\mathbb R^n$ is a distributive lattice with respect to this ordering. Let us consider lattices $L\subseteq \mathbb R^n$ which are lattices with respect to the induced partial order, and which are closed under translations:

$$\forall_{x\in L}\forall_{t\in\mathbb R}\quad x+t\in L$$

where $(x+t)_k:= x_k+t$ for every $k=1\ldots n$.

Call such lattices $L$ translational lattices in $\mathbb R^n$.

REMARK In general, translational lattices in $\mathbb R^n$ are NOT sublattices of $\mathbb R^n$; they are often not distributive, and that's the point. We are searching for a modular non-distributive lattice which satisfies an additional "polynomial" identity not satisfied by some modular lattices.

The goal of a full classification of translation lattices in all spaces $\mathbb R^n$ is realistic. The topic is a mix of combinatorics and geometry (may be with a touch of topology, next to nothing). The translational lattices have rather simple geometric shapes, possible to classify.

DIGRESSION (sorry for my ignorance) Were any other equational classes of lattices studied besides the modular lattices, distributive lattices, and all lattices?

Can a composition with itself of a universal self-map be non-universal?

2013-05-12T20:14:40Z

I have formulated (and published) the notion of a universal map (and of a universal morphism), and the problems below, in the early 1960-ies.

DEFINITION A continuous map $u: X\rightarrow Y$ is called universal $\Leftarrow:\Rightarrow$ for every continuous map $f:X\rightarrow Y$ there exists a point $p\in X$ such that $f(p)=u(p)$.

Questions:

Does there exist a topological space $X$, and a universal map $u:X\rightarrow X$ such that $u\circ u : X\rightarrow X$ is not universal?
Can $X$ in the question above be a 2-dimensional finite polyhedron?

I am convinced that the answer to the first question is yes. Most likely $X$ can be a finite polyhedron. I have formulated also a categorical version of this question, and a more special for monoids too (a monoid is a category with just one object after all). The one for monoids was solved by Ralph N McKenzie in May of 2006 (private communication). He considered the free type of a construction. There should be a natural way to topologize Mckenzie's example or any monoid example, i.e. to get a topological space with the desired properties out of the respective monoid. But to obtain a polyhedron is another matter.

Within topology the theory of universal maps includes the topological dimension theory and the theory of the fixed point property; it is connected to the theory of manifolds, and to the stable cohomotopy groups. Couniversal morphisms would be perhaps of interest to the theory of Banach Algebras. While applying a version of the fixed point property in functional analysis and to differential equations may require an odd introductory step, it may be more natural to apply universal maps (in at least one case it is indeed).

EXAMPLE (late 1960ies) The composition of two universal mappings between 2-dimensional polyhedra does not have to be universal. Indeed, let $M$ be the Möbius strip (compact, with its boundary). Let $f : M \rightarrow D$ be the map corresponding to the identification of the equator points, so that it sends the equator to the center of the unit complex disk $D$. Let $g : D\rightarrow D$ be given by $\forall_{z\in D}\ g(z) := z^2$. Then $f\ \ g$ are universal, while $g\circ f$ is not.

Justification: all three maps $f\ \ g\ \ \ g\circ f$ are into disk $D$. The universality of the first two, and the non-universality of the composition, follows from the following elementary characterization of the universal mappings into a Euclidean ball (or cube):

Let $B$ be a closed ball in an $n$-dimensional Euclidean space. Let $S$ be the boundary of $B$. Consider an arbitrary continuous function $f:X\rightarrow B$, where $X$ is an arbitrary topological space, and let $F := f^{-1}(S)$. Then the following two properties of $f$ are equivalent:

$f$ is universal;
there does not exist a continuous function $g:X\rightarrow B$ such that $g(X)\subseteq S$ and $g|F=f|F.$

For instance, let's go back to the map $f:M\rightarrow D$ of the above example. We can start with a definition of the Mobius strip $M$ as a topological quotient of the following subspace $M'$ of the complex plane $\mathbb C$:

$$M'\ :=\ \{z\in\mathbb C: \frac 12\le |z|\le 1\}$$

Space $M$ (the Mobius strip) is obtained from $M'$ by identifying all pairs of points $z\ \ w\ \in\ M'$ such that $w=-z$ and $|z|=|w|=\frac 12$. Now $f :M \rightarrow D$ is induced by $\phi:M'\rightarrow D$ given by:

$$\forall_{z\in M'}\quad \phi(z) := (2\cdot |z|-1)\cdot z$$

Answer by Wlodzimierz Holsztynski for Simplex with edges of length at least s having smallest circumradius

2013-05-15T03:36:58Z

Let $H$ be a Hilbert space. Let $e_1\ \ldots\ e_n$ be vectors such that $\forall_{k=1\ldots n}\ \|e_k\|\le 1$ for a natural $n$ (to me natural numbers are always positive). I am addressing $(n-1)$-simplex for the sake of avoiding eyesores. Then:

$$ 0\ \le\ (e_1+\ldots +e_n)^2\ \ \le\ \ n\ +\ 2\cdot\sum_{j\ne k}e_j\cdot e_k $$

It follows that there exist $j\ne k$ such that

$$ e_j\cdot e_k\ \ \ge\ \ \frac{-n}{2\cdot\binom n2}\ =\ \frac1{1-n}$$

and

$$\|e_j-e_k\|\ =\ \sqrt{e_j^2 - 2\cdot e_j\cdot e_k + e_k^2}\ \le\ \sqrt{2\cdot(1+\frac 1{n-1})}\ =\ \sqrt{\frac{2\cdot n}{n-1}}$$

i.e.

$$\|e_j-e_k\|\ \le\ \sqrt{\frac{2\cdot n}{n-1}}$$

REMARK 0 We get equality above $\Leftrightarrow$ $\|e_j\|=\|e_k\|=1$ and the $\binom n2$ values $e_j\cdot e_k$ are all equal one to another, i.e. all distances $\|e_j-e_k\|$ are all equal. (Then of course all dot products are equal to $\frac 1{1-n}$, and the distances to $\sqrt{\frac{2\cdot n}{n-1}}$).

Now let's reverse our point of view. Let $e_1\ldots e_k\in H$ be such that $\|e_j-e_k\|\ge s$ for every $j\ k = 1\ldots n$ such that $j\ne k$, where $s>0$ is an arbitrary positive real constant. Furthermore, we may assume that the center of a sphere which has $e_1\ldots e_k$ for its points coincides with the origin of $H$, so that $\|e_1\|=\ldots =\|e_n|=r$ for certain positive real $r$. Then

$$ s\ \ \le\ \ r\cdot \sqrt{\frac{2\cdot n}{n-1}}$$

i.e.

$$ r\ \ \ge\ \ s\cdot\sqrt{\frac{n-1}{2\cdot n}}$$

The equality holds $\Leftrightarrow$ all edges have the same length $\|e_j-e_k\|=s$, and vectors $e_1\ldots e_n$ have to belong to a common $(n-1)$-dimensional linear subspace of $H$ (substitute linear by affine in the general case).

Answer by Wlodzimierz Holsztynski for Picturing a Certain Torus and Klein Bottle

2013-05-15T04:20:37Z

(I tried THREE times to post my answer as a comment. It's too hard. Let me do it here.)

@ARupiński, you must be fighting imaginary demons, because there is not much to visualize, this question is simple (not research :-). Let $S$ be the complex unit circle. First consider the Klein bottle: $K2\times\{1\}\subseteq K_2\times S$. Just agree that the normal vector to $(p\ 1)$ is $(p\ i)$ for every $p\in K_2$. As you see (it's trivial) you got a smooth field of normal non-zero vectors for your entire Klein bottle $K2\times\{1\}$. That's one side of your Klein bottle. The other side is given by normal vectors $(p\ −i)$. Your Klein bottle copy is $2$−sided. (Sorry for the imperfect notation for the normal vectors).

Now the challenge is not in seeing things but in writing them down (I wish I could do it without too much pain). Let me start with a convenient (for this text) definition of the Klein bottle. Let $\mathbb C$ be the complex plane. Let

$$M'\ \ :=\ \ \{z\in \mathbb C\ :\ \frac 12\le |z|\le 1\}$$

Identify $z\ \ w\in M'\ \Leftrightarrow\ \ w=-z$ and $|z|=|w|=\frac 12$. The result is a Mobius strip, call it $M$. That's one half of your Klein bottle. The boundary of $M$ is the unit circle. Points obtained from the identification of complex numbers $z$ of module $\frac 12$ form the so called equator $E$ of $M$. The whole Klein bottle $K_2$ is the so-called double of $M$. We only need to know at this point that in the topological space $K_2$ equator $E$ is contained in the topological interior of $M$.

Consider the torus $T := E\times S\subseteq K_2\times S$. At each point $([z]\ 1)\in E\times S\subseteq K_2\times S$, where $|z|= \frac 12$ (i.e. $[z]\in E$) consider two normal vectors: $(z\ 0)$ and $-(z\ 0)$ -- these two vectors are on the locally opposite sides of the torus in the $K_2\times S$ &nbs[; space. Let's move one of the said vectors around, say vector $(\frac 12\ 0)$ starting at $([\frac 12]\ 0)\in E\times S$. Moving in our vector field along the equator we will get back to the same point $([\frac 12]\ 1)$ of the equator, but we will end up this time with the opposite vector, namely with vector $-(\frac 12\ 0)$. This shows that the locally different sides are the same globally.

REMARK Sorry to do such a lousy job of explaining. If you feel that this mini-exposition is worthy of improving, and if you feel like improving it, then please just go ahead and edit this text.

A characterization of Hilbert spaces?

2013-05-10T22:39:33Z

My question was prompted by an earlier MO by @Daniel:

Duality map in strictly convex Banach spaces

I will even use his symbol $\phi$ below.

Let $B$ be an arbitrary Banach space. Let $S := \{x\in B:\|x\|=1\}$ be its unit sphere. Let $\Gamma := \{f\in B^*: \|f\|=1\}$ be the unit sphere in the dual space $B^*$.

QUESTION Are the following two conditions on $B$ equivalent:

B is isometric to a Hilbert space.
There exists an isometry $\phi: \Gamma \rightarrow S$ such that $\forall_{f\in\Gamma}\ f(\phi(f))=1$.

The finite-dimensional case is especially basic.

REMARK 0 Perhaps similar questions were asked in the past (on MO too?)--please, let me know.

REMARK 2 The case of $\mathbb R^2$ and its two dual but isometric norms $L_\infty\quad L_1$ is interesting. The general question related to the one above is to describe all Banach spaces which are isometric to their dual space. Is there any beside the Hilbert spaces and $\mathbb R^2$ with the norm(s) just mentioned above?

Answer by Wlodzimierz Holsztynski for Awfully sophisticated proof for simple facts

2013-05-11T04:02:26Z

Around year 1970 a popular way to compute cohomology groups of the finite cyclic groups was by applying spectral sequences (which was quite an overkill).

Answer by Wlodzimierz Holsztynski for Duality map in strictly convex Banach spaces

2013-05-10T05:08:29Z

There are $n$-dim Banach spaces for which $\phi$ is not Lipschitz in every dimension $n\ge 2$.

It is enough to provide an example in dimension $2$. The higher dimensional examples are obtained by rotating the lower dimensional examples.

In dimension $2$ the required norm, in $\mathbb R^2$, can be given by:

$$\|(x\ y)\| := (x^4+y^4)^{\frac 14}$$

(I am willing to provide the details if asked to--it's a simple matter).

Answer by Wlodzimierz Holsztynski for Axiom of Choice and Continuous function

2013-02-18T19:22:21Z

Look Ma, no axiom of choice!

THEOREM 0 Let $X$ be a compact space. Let $\Phi$ be a non-empty family of closed subsets of $X$, $F := \bigcap \Phi$, and $G\supseteq F$ an open subset of $X$. Then there exists a finite $\Phi_0\subseteq \Phi$ such that $\bigcap\Phi_0\subseteq G$.

PROOF Family $\Gamma\ :=\ G\cup\{X\setminus A : A \in \Phi\}$ is an open covering of $X$. Etc. END of PROOF

As an instant corollary we get:

THEOREM 1 Let $X$ be a compact space. Let $\Phi$ be a non-empty family of closed subsets of $X$, $F := \bigcap \Phi$, and $G\supseteq F$ an open subset of $X$. Assume also that $\Phi$ is linearly ordered by $\subseteq$. Then there exists $A\in \Phi$ such that $A \subseteq G$.

THEOREM 2 Let $(X\ d)$ be a compact metric space. Let $W$ be an open subset of $X^2$, such that $\Delta_X\subset W$, where $\Delta_X := \{(x\ x):x\in X\}$. Then there exists $\delta > 0$ such that $V_X(\delta)\subseteq W$, where $V_X(\delta) := \{(x\ y)\in X^2 : d(x\ y)\le\delta\}$.

PROOF Apply Theorem 1 to $X^2$ as a replacement of $X$ of Theorem 1; etc. END of PROOF

THEOREM 3 Let $f : X\rightarrow Y$ be an arbitrary continuous function of a metric compact space $(X\ d_X)$ into an arbitrary metric space $(Y\ d_Y)$. Then function $f$ is uniformly continuous.

PROOF Let $\epsilon > 0$. Let $W\subseteq X^2$ be the inverse image of $V_Y(\epsilon)$ under function $f\times f$. There exists, by Theorem 2, $\delta > 0$ such that $V_X(\delta)\subseteq W$. Then $d_Y(f(x')\ f(x''))\le\epsilon$ for every $x'\ x''\in X$ such that $d_X(x'\ x'')\le\delta$. END of PROOF

Answer by Wlodzimierz Holsztynski for Mathematical habits of thought and action which would be of use to non-mathematicians

2013-05-03T18:18:15Z

An applied problem is often formulated as follows:

Go 1 mile East, then 1 mile North, then 1 mile West. You'll get to your destination D, and that's your goal.

And they follow the direction like slaves, they go East-North-West. They can't tell the difference between the goal and the description of the goal. It takes sometimes a mathematician to tell them to go North right away.

Answer by Wlodzimierz Holsztynski for Pathological Examples of Dimension

2013-05-03T04:42:50Z

If one of the two topological spaces $X\ \ Y$ is compact and $1$-dimensional then the logarithmic equality for the covering dimension holds:

$$\dim(X\times Y)\ \ =\ \ \dim(X) + \dim(Y)$$

Thus in the compact case Erdős example cannot be matched, one needs to work with the dimensions greater or equal $2$. This was accomplished by Pontryagin, who provided continua $X\ Y$ of dimension $2$ such that their product $X\times Y$ was $3$-dimensional.

Next, a more subtle example was given by Boltyansky. His 2-dimensional continuum $B$ was such that its square was 3-dimensional $\dim(B^2) = 3$.

These examples can be understood (better) from the point of view of homological (or cohomological) dimension, in the combination with the simple underlying geometric nature of these examples.

Answer by Wlodzimierz Holsztynski for Pathological Examples of Dimension

2013-05-03T04:27:17Z

@Chris: the Cech homology of any compact $n$-dimensional space $X$ (the covering dimension is meant here) vanishes in the dimensions above $n$ because:

$X$ is an inverse limit of $n$-dimensional finite polyhedra;
Cech homology of compact spaces is continuous (with respect to the inverse limit operation).

It follows that Cech homology of any Hawaiian $n$-dimensional earring vanishes in every dimension above $n$.

Answer by Wlodzimierz Holsztynski for $n$-in-a-row game on $\mathbb{R}^2$

2013-05-02T03:31:05Z

Let's consider a version which is harder on the first player, when s/he needs the whole straight line, not just a segment, to win. S/he can still win in $7$ moves. I'll present a brute force solution (strategy).

We may assume that the first player chooses moves $A_1\ A_2\ldots$ without any two pairs belonging to two different parallel lines, and still will follow the strategy described below (it'd be simpler to consider the projective plane :-). Furthermore, we may also assume that no three different points of the first $5$ moves, $A_1\ \ldots\ A_5$, are collinear.

Strategy:

$A_2$ does not belong to the straight line $L(A_1\ B_1)$; thus $B_1\notin L(A_1\ A_2)$;
none of the points $B_1\ B_2$ belongs to $L(A_1\ A_3)\cup L(A_2\ A_3)$;
none of the points $B_1\ B_2\ B_3$ belongs to $L(A_i\ A_3)\cup L(A_i\ A_4)\cup L(A_3\ A_4)$, for at least one value $i \in \{1\ 2\}$;
none of the points $B_1\ B_2\ B_3\ B_4$ belongs to $L(A_i\ A_5)\cup L(A_3\ A_5)\cup L(A_4\ A_5)$, for the same $i$ as before;
there are $3$ partitions of $\{A_i\ A_3\ A_4\ A_5\}$ into disjoint pairs of points; each pair of pairs produces an intersection point of the lines which pass through each pair; of points $B_1\ \ldots\ B_5$ only points $B_4\ B_5$ can belong to the union of lines which pass through any two points $A_i\ A_3\ A_4\ A_5$ (and $B_4$ belongs to at the most one of these lines); moreover--point $B_5$ does not belong to any two lines from different partitions (only points of the first player do); thus the lines of one of the mentioned partitions into disjoint pairs are free from all points $B_1\ \ldots\ B_5$; let the first player choose the intersection point of this pair of lines as $A_6$;
point $B_6$ can belong to at the most one line just mentioned above; thus the first player can choose a point $A_7$ on the other line, and will obtain $4$ points on this line, while the second player has none.

CONCLUSION: for $n=4$ the first player can win in $7$ moves.

REMARK We considered a total of $13$ points $A_1\ B_1\ \ldots\ B_6\ A_7$, which is the number of points of the projective plane over field $\mathit {GF}(3)$. It must be simple to decide if the above argument carries over to the finite $13$-point projective plane but I am too tired of mathoverflow $\LaTeX$ to see anything anymore.

Answer by Wlodzimierz Holsztynski for Is there an algebraic approach to metric spaces?

2013-04-07T04:18:25Z

An algebraic context for metric spaces--attractive to me--can be be provided by translation lattices, introduced (in the bounded, distributive case) by Irving Kaplansky, Lattices of continuous functions II, Amer. J. Math. 70 (1948), pp. 626-634; see also Włodzimierz Holsztyński, Lattices with real numbers as additive operators, Dissertationes Mathematicae LXII, Warszawa 1969.

Answer by Wlodzimierz Holsztynski for One tetrahedron inside another tetrahedron

2013-02-17T06:39:35Z

The conjecture about the quotient of the sums of the lengths of edges for a tetrahedron inside another was formulated (a long time ago) by Jerzy Browkin, who saw that, somewhat paradoxically it can be greater than 1, and even arbitrarily close to 4/3, as his examples proved it. He asked if $\frac 43$ is the least upper bound of such quotients.

The conjecture was solved in 1961 by Wlodzimierz Kuperberg and myself. We published it in 1962 or 1963, in Polish, in Wiadomosci Matematyczne (of Polish Mathematical Society). Years later an English translation was published in Alabama Journal of Mathematics (or similar). And a powerful complete multidimensional generalization was obtained and published by Carl Linderholm, An inequality for simplices .

Ancient Greeks could easily formulate and solve the given problem (in 3d) but (as long as we know) never did--it took Jerzy Browkin to raise it.

Answer by Wlodzimierz Holsztynski for Bounded Hamming distance

2013-04-30T08:35:28Z

It seems, especially in the light of @Yuichiro answer, that it makes sense for me to share here with you my original constant weight codes, which I have discovered in 1977-78 but just absolutely couldn't believe that they were not known. Only years later I got clear evidence that they were still unknown to the public (only then I posted them on pl.sci.matematyka, and I informed about them Neil Sloan and his co-maintainer of the ECC tables--both worked at Bell Labs at the time; I didn't get any feedback from them though).

My construction mostly doesn't care about the finiteness. Let $K$ be an arbitrary field, let $L$ be an arbitrary proper subfield of $K$. Let $P(K)\ \ P(L)$ be their projective lines (1-dim projective spaces); we may assume that $P(L)\subset P(K)$ -- it's a harmless abuse. Let $G\ H$ be the projective groups of $P(K)\ P(L)$ respectively. Let $\Gamma := \Gamma(K\ L)$ be the family of all images of $P(L)$ under the projective maps from $G$:

$$\Gamma := \{f(P(L)) : f\in P(K)\}$$

That's it. We may call the members of $\Gamma$ to be circles. For every three different points $x\ y\ z\in P(K)$ there exists exactly one circle which contains all three of them. When $K$ is a finite field then $\Gamma$ is the best possible (even perfect or similar terminology) constant-weight code--instead of considering the binary sequences we deal equivalently with subsets of $P(K)$, or here simply with circles.

Let's say that $|K|=p^n$ and $|L|=p^m$, where $p$ is a prime, and $0 < m < n$ are two natural numbers. Thus $w:=p^m+1$, while @sams' $n$ is $p^n+1$ here (sorry for that). The minimal distance between the codes is $a := 2\cdot (p^m-1)$. And that's what is important for the standard theory, while the maximal distance between the codes is $b:=2\cdot (p^m+1) = a+4$.

Of course $$|\Gamma|\ \ =\ \ \frac{\binom \nu 3}{\binom \mu 3}$$ where $$\nu := p^n+1\qquad\quad \mu=p^m+1$$

The property of circles (exactly one circle passing through any three different points) follows from the exact 3-transitivity of $G$, and the identification $H\subset G$.

Edit: I rushed. Of course $p^n$ must be a power of $p^m$, i.e. $n$ must be a multiple of $m$.

Answer by Wlodzimierz Holsztynski for Existence of a continuous and unbounded map $f$ with $f(f(x))=x$

2013-04-30T05:11:12Z

Let's assume the $\ell^2$-norm topology. Isn't it true that there exists a homeomorphism $$h :\ H\rightarrow H\setminus\{0\}$$ which is an identity when restricted to the unit sphere and outside the unit ball? Then the rest is straightforward.

Indeed, first of all consider involution $i : H\setminus\{0\}\rightarrow H\setminus\{0\}$ given by formula

$$i(x) := -\frac x{x^2}$$

Now the desired involution $j : H\rightarrow H$ is given by:

$$ j := h^{-1}\circ i\circ h$$

Regards,
Wlod

Answer by Wlodzimierz Holsztynski for A question on metrizable space

2013-02-24T05:30:53Z

Answer to Q1:

Let $(X\ d)$ be a metric space. I call $A\subseteq X$ $\epsilon$-dispersed $\quad\Leftarrow:\Rightarrow\quad\forall_{x\ y\in A}\ \left(\left(x\ne y\right)\Rightarrow d(x\ y)\ge \epsilon\right)$.

Let $A_\epsilon$ be a maximal $\epsilon$-dispersed set in $(X\ d)$ for every $\epsilon > 0$ (apply Kuratowski-Zorn theorem). Then $\bigcup_{n=1}^\infty\ A_{\frac 1n}$ is dense in $(X\ d)$. (The rest is obvious).

Answer to Q2:

(I don't see any use for $\omega_1$--am I wrong?)

I call a topological space singular $\quad\Leftarrow:\Rightarrow\quad$ it has exactly one limit point (i.e. non-isolated).

THEOREM The topological product of an arbitrary Lindelöf space by an arbitrary singular Lindelöf space is Lindelöf.

PROOF In arbitrary singular Lindelöf space the complement of any open set, which contains the limit point, is countable. The rest is obvious.

Answer by Wlodzimierz Holsztynski for Which topological spaces are (topological) groups?

2013-04-29T16:41:09Z

A necessary condition for a Hausdorff compact space to admit the structure of a topological group is the Suslin condition (I hope I am using proper terminology)

every family of pair-wise disjoint open sets is countable.

This is so because Hausdorff compact topological groups admit Haar measure.

Answer by Wlodzimierz Holsztynski for Hamming codes from overlapping vectors

2013-04-28T20:39:46Z

Special case. There are wonderful (I am very sorry, I don't remember the name of their discoverer) binary sequences $(x_n)_{n\in \mathbb Z}$, which have period $2^m$, and such that the consecutive $2^m$ subvectors of length $m$ are all different (hence they exhaust all binary vectors of length $m$). Thus you may consider the induced sequence of length $n:=2^m$, with indices ordered cyclically, or you may consider an ordinary subvector of the length equal to $n:=2^m + m -1$. For such special $n$ your question now is reduced to the most fundamental question of the theory of the error correcting codes, where you simply (:-) ask about the maximal size of codes of length $m$ and distance $d$.

Answer by Wlodzimierz Holsztynski for 3-coloring of specific planar graphs

2013-04-21T01:58:04Z

The complete graph $K_4$ (the graph of the edges of a tetrahedron) is a counter-example, it requires exactly $4$ colors, no less.

Answer by Wlodzimierz Holsztynski for splitting one limit into two?

2013-04-08T07:24:14Z

The first (single) limit is totally blind to terms $a_{k\ n}\ \ b_{k\ n}$ for all $k > n$, while the second (double) limit depends on them. Thus the two limits are hardly related at all.

In other words, the single limit considers finite segments, and the double limit the infinite segments. To compare these two there should be perhaps a relation given between the finite and infinite sums.

Also, something should be said about denominators staying reasonably away from $0$; well--something :-)

(I am surprised that the m-th sums in the single limit case have exactly m terms--I'd expect a more flexible situation).

Answer by Wlodzimierz Holsztynski for Can Lipschitz maps increase the Lebesgue dimension ?

2013-04-06T04:56:08Z

Let me add to the examples by Benoit and Victor another Cantor example, this time a straightforward naive one rather than ingenious.

Consider at and just after stage $0$ a closed interval $I$ with the standard (Euclidean) metrics but of length $\frac{11}{10}$. At stage $k>0$ remove the center open interval of length $\frac 1{2^{k-1}\times 11^k}$ of each interval left after the previous stage $k-1$. After all stages $0\ 1\ \ldots$, in the remaining set $C$ in addition to the Euclidean metrics consider also the following pseudo-metrics:

$$d(x\ y) = |x-y| - s_{x\ y}$$

where $s_{x\ y}$ is the sum of the lengths of all removed intervals which are between points $x\ y$. The identity map from Euclidean $C$ to $C$ with the pseudo-metric $d$ is Lipschitz with constant 1. Let $C'$ be the metric space induced by $C$. Then $C'$ is homeomorphic to a nondegenerated closed interval, and the map induced by the identity on $C$ is Lipschitz with constant 1.

Actually, C' is isometric with the unit Euclidean interval $[0;1]$.

Answer by Wlodzimierz Holsztynski for Every continuous function is homotopic to a locally Lipschitz one

2013-04-06T03:52:38Z

A modest start.

Consider two finite geometric simplicial complexes with reasonable metrics, e.g. inherited from the ambient Euclidean (or Banach) space (where simplices are affine). Then every continuous function $f$ between them is uniformly approximated by the simplicial maps of iterated baricentric subdivisions of the first complex into the second complex. All these simplicial maps are Lipschitz. When approximation is close enough to $f$ then it is homotopically equivalent to $f$. This gives a positive answer to your question for finite geometric simplicial complexes.

REMARK 0 For the sake of obtaining a Lipschitz map homotopic to a given continuous map one does not need to subdivide the second complex.

On the other hand, it is not difficult to provide two metric functions (distance functions) for the unit circle $S^1$ (thus let's talk about two metric spaces anyway) such that the identity map from one of them to another is not homotopic to any locally Lipschitz function. Indeed, there will not exist any locally Lipschitz function at all (not even at any inverse image of any non-empty open set) from the first space onto the second one (under the fixed but properly selected metric functions; the first one can be the standard metrics).

REMARK 1 Instead of $S^1$ we could consider a space consisting of a convergent sequence and its limit, endowed with two distance functions such that the identity is not Lipschitz (at the limit point). The only map homotopic to the identity is the identity, hence another instance of the negative answer. But $S^1$ is nicer :-)

Answer by Wlodzimierz Holsztynski for Combinatorial distance between simplicial complexes

2013-04-06T01:59:24Z

Am I missing something? You need to remove what you need to remove, and to add what you need to add.

Thus the first stage: remove one of the highest dimensional simplices from the first complex which does not belong to the second complex. Do it again and again until there is nothing more to do.

The second and last stage: add one of the lowest dimension simplices which is missing in your construction but is present in the second complex. Do it again and again until there is nothing more to do.

That's it. If the weight of each simplex is simply $1$ then the distance between two simplicial combinatorial complexes is equal to the cardinality of the symmetric difference of the given two complexes (as was already mentioned).

Sorry if I am crudely off.

Answer by Wlodzimierz Holsztynski for Can you explicitly write R^2 as a disjoint union of two totally path disconnected sets?

2013-02-24T21:01:35Z

I'll apply the following simple result:

THEOREM Let $f : I\rightarrow X$ be an arbitrary non-constant continuous function (a path) of interval $I:=[0;1]$ into an arbitrary topological space $X$. Then there exist continuous maps $\alpha:I\rightarrow I$ and $g:I\rightarrow X$ such that $f = g\circ \alpha$, and $g$ is not constant on any non-empty open subinterval of $I$.

Here is a simple positive solution for the question of this thread, and proof:

Let $\mathbb C := \mathbb R^2$ be the complex plane. Let $K \subseteq \mathbb C$ be a Knaster pseudo-arc. Let $$L := i\cdot K := \{i\cdot z : z \in K\}$$

where $i^2=-1$. Let $D$ be a dense countable subset of $\mathbb C$. Define $$A := \left(\bigcup_{d\in D}\ \left(d+K\right)\right)\cup\left(\bigcup_{d\in D}\ \left(d+L\right)\right)$$

where $d+X := \{d+x:x\in X\}$. Finally, let $$B := \mathbb C\setminus A$$

Then $\dim(B) = 0$, and $B$ does not contain any non-constant path.

Also, there does not exist any non-constant continuous map $f : I \rightarrow A$ --indeed, if there was one then we may assume that it is not constant on any open subinterval of $I$. Then the inverse images:

$(\bigcirc^{-1}f)(d+K)\quad$ and $\quad(\bigcirc^{-1}f)(d+L)$

would be 0-dimensional closed subsets of $I$, for every $d\in D$. Thus $I$ would be a countable union of 0-dimensional closed subsets, which is a contradiction. It means that $A$ does not contain any image of any non-constant path.

This completes a positive answer to the Question of this thread.

Answer by Wlodzimierz Holsztynski for Cech cohomology as a colimit over maps to a CW complex

2013-02-25T05:31:26Z

A short answer for now (a longer edit or answer perhaps later or perhaps its not needed).

When you use continuous functions then your category for a compact space $X$ is virtually just the space $X$ itself. In the given context it does not buy much. But when you use homotopy classes instead then you get the shape of compact $X$ of the continuous shape theory (as in the natural model of the axiomatic shape theory). The continuous shape functor is like a Dedekind section between shape functors and continuous homotopy invariant functors (like Cech homology and cohomology, etc)--these latter ones are automatically shape invariant. Topological category of compact spaces is like Earth, and the continuous shape functor is like a cosmic station for all these other "cosmic" functors.

Comment by Wlodzimierz Holsztynski

2013-05-20T07:25:49Z

@Heng, I'd say hit the textbooks or Google (perhaps wikipedia, I don't know).

Comment by Wlodzimierz Holsztynski

2013-05-20T05:20:55Z

It took me only a dozen editions to get $\LaTeX$ right above. It still doesn't look pretty but I am not going to beautify it :-)

Comment by Wlodzimierz Holsztynski

2013-05-20T05:18:02Z

@mbsq: I'll guess and signal a direction toward an answer (I don't know your formalism): $\{f|f:\omega\rightarrow \omega_1\} =\bigcup_{\alpha < \omega_1}\{i_{\alpha\omega_1}\circ(f:\omega\rightarrow\alpha)\}$. (you and OM are killing me with $\LaTeX$. Yes, this is all just a matter of convention--said Alice in Wonderland.

Comment by Wlodzimierz Holsztynski

2013-05-19T19:26:49Z

@Qfwfq, there was plenty of (objective) evidence during my years. I'll expand my "Answer" above to reflect it somewhat. -o=o- @mbsq, what you call so cheerfully "absurdity" is the mathematical reality at least since Eilenberg + (Mac Lane, Cartan, Steenrod, Grothendieck, ...), while I am old enough to remember the good set-theoretical, pre-category times.

Comment by Wlodzimierz Holsztynski

2013-05-19T00:56:13Z

Isn't thee theorem about separability of the product of continuum many separable spaces due to Edward Szpilrajn-Marczewski? BTW, it holds for the product of $2^a$ spaces with dense subsets of cardinality less or equal $a$.

Comment by Wlodzimierz Holsztynski

2013-05-19T00:16:53Z

@Dustin, I see only a very weak analogy between the 2d and 3d situations described in your question. What am I missing?

Comment by Wlodzimierz Holsztynski

2013-05-17T01:32:03Z

Thank you Gerhard "web traveler" Paseman.

Comment by Wlodzimierz Holsztynski

2013-05-16T18:36:18Z

C'mon Andreas, "Balcerzyk" is not such a hard name :-)

Comment by Wlodzimierz Holsztynski

2013-05-16T18:07:52Z

I am curious: what is what?

Comment by Wlodzimierz Holsztynski

2013-05-16T08:51:24Z

Wow! (Just unbelievable :-).

Comment by Wlodzimierz Holsztynski

2013-05-16T00:02:53Z

Perhaps someone may fix the typo in (*).

Comment by Wlodzimierz Holsztynski

2013-05-15T23:39:49Z

@Sergei, your theorem/proof is very-very nice, you had an idea after an idea.

Comment by Wlodzimierz Holsztynski

2013-05-15T07:59:12Z

@'9999', I am curious--why topology tag?

Comment by Wlodzimierz Holsztynski

2013-05-15T07:57:26Z

"You can use the definition of Riemann" -- @'9999', that's very generous of you.

Comment by Wlodzimierz Holsztynski

2013-05-15T07:13:01Z

@Sergei, somehow I missed your answer, I discovered your answer just a moment ago, I am really sorry. I'll carefully read your answer within 24h, and will comment on it (even if I will not get it :-), but at least I will admit it).